Consider the integral functional $$J(y)=\int_{-1}^{1}y^2(x)(1-y'(x))^2dx$$ defined on the domain $$D=\{y\in C^1[-1,1]|y(-1)=0,y(1)=1\}$$ I'm asked to show given any $\epsilon>0$, there always exists a $y\in D$ such that $|J(y)|<\epsilon$.
Currently I have found a continuous function $y_0$ that satisfies $J(y_0)=0$ is given by $$y_0(x)=\begin{cases}0&x\in[-1,0]\\x&x\in[0,1]\end{cases}$$ Is that possible to finish the proof by using this observation and proving the continuity of $J$?
Your $y_0$ can be plugged into the integral formula for $J$, since $y'$ is defined everywhere except $0$, and you get $J(y_0)=0$ formally. However $y_0$ is not in $D$ since it is not $C^1$.
To deal with that along the lines of your suggestion, change the behavior near $x=0$ (e.g. on $[0,\delta]$) so you have a $C^1$ function. This means you want a function that is $0$ at $x=0$ and has derivative $0$ at $x=0$ and is $\delta$ at $x=\delta$ and has derivative $1$ at $x=\delta$. For each fixed $\delta$, there is a cubic polynomial with all four of those properties. You might hope that the coefficients are such that this polynomial (extended as you suggested off $[0,\delta]$) yields a small value of $J$ for small $\delta$...which you can check by an explicit if tedious calculation.
Normally people use an analytical tool called a mollifier to avoid this kind of explicit calculation, but here the explicit calculation is straightforward enough.